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Showing that [G:H]=[G:K][K:H]

  • #1
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Homework Statement


Prove that if ##H\leq K\leq G## and ##[G : K]## and ##[K : H]## are finite, then ##[G:K]\cdot [K : H]=[G : H]##

Homework Equations




The Attempt at a Solution


Here is my attempt. We first note that $$G = \bigcup_{x\in G} xK ~\text{and}~ K=\bigcup_{y\in K}yH$$.
So $$G = \bigcup_{x\in G} x\bigcup_{y\in K}yH = \bigcup_{(x,y)\in G \times K}xyH$$ Now we show that ##G## is a disjoint union. Suppose that ##xyH \cap x'y'H \not = \emptyset##. Then there is some nontrivial ##x## such that ##x = xyh=x'y'h'## for some ##h,h'\in H##. But then it's clear that ##xyH=x'y'H##. Hence we have a disjoint union.

This is where I get stuck. Is there some way I can glean from what I proved the fact that ##[G:K]\cdot [K : H]=[G : H]##?
 

Answers and Replies

  • #2
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9,146
So $$G = \bigcup_{x\in G} x\bigcup_{y\in K}yH = \bigcup_{(x,y)\in G \times K}xyH$$
I'd say you have ##G= \cup_{(xy)\in G}\,(xy)H##, because there is no need for an external direct product. All ##y\in K## are also in ##G## and so are ##xy \in G##. No need for a separation.
Now we show that ##G## is a disjoint union. Suppose that ##xyH \cap x'y'H \not = \emptyset##. Then there is some nontrivial ##x##...
Say we have a ##g\in G##, regardless of whether it is ##g=e##.
... such that ##x = xyh=x'y'h'##...
##g=xyh=x'y'h'##
... for some ##h,h'\in H##. But then it's clear that ##xyH=x'y'H##. Hence we have a disjoint union.
I do not see it immediately, since ##x## and ##y## do not commute. IMO there should be some more reasoning, especially as the double usage of ##x## in one and the same equation is a bit weird.
This is where I get stuck. Is there some way I can glean from what I proved the fact that ##[G:K]\cdot [K : H]=[G : H]##?
See my first remark. The introduction of an artificial external direct product, let it be groups or just sets, is the reason why you are trapped.
 
  • #3
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44
I'd say you have ##G= \cup_{(xy)\in G}\,(xy)H##, because there is no need for an external direct product. All ##y\in K## are also in ##G## and so are ##xy \in G##. No need for a separation.
Say we have a ##g\in G##, regardless of whether it is ##g=e##.##g=xyh=x'y'h'##
I do not see it immediately, since ##x## and ##y## do not commute. IMO there should be some more reasoning, especially as the double usage of ##x## in one and the same equation is a bit weird.

See my first remark. The introduction of an artificial external direct product, let it be groups or just sets, is the reason why you are trapped.
Well even if I have the disjoint union ##G = \bigcup_{(xy)\in G} xyH## I'm not seeing how to conclude that ##[G: K][K : H] = [G : H]##
 
  • #4
12,633
9,146
You could show ##|G|\stackrel{(*)}{=}|G:U|\cdot |U|## instead, and simply use this equation repeatedly. Your approach would do the same, but with all versions of the proof for ##(*)##, i.e. all steps for ##H\leq K\; , \;H\leq G\; , \;K\leq G## repeated in three versions, if you count the elements in your cosets and the number of cosets.
 

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